Question

Pleaseee helppp with thisss asappp

Answer 1

We are given with a equation of Circle and we need to find it's radius and it's equation in standard form . But , let's recall , the standard equation of a circle is ${\bf (x-h)^{2}+(y-k)^{2}=r^{2}}$ where (h,k) is the centre of the circle and radius is r . Proceeding further ;

${:\implies \quad \sf x^{2}+12y+22x+y^{2}-167=0}$

Collecting x terms , y terms and transposing the constant to RHS ;

${:\implies \quad \sf (x^{2}+22x)+(y^{2}+12y)=167}$

Now , as in standard equation their is a whole square , so we need to develop a whole square in LHS , for which we will use completing the square method , as coefficient of x² and y² is 1 , so adding 121 and 36 to LHS and RHS .

${:\implies \quad \sf (x^{2}+22x+121)+(y^{2}+12y+36)=167+121+36}$

${:\implies \quad \sf (x+11)^{2}+(y+6)^{2}=324\quad \qquad \{\because a^{2}+2ab+b^{2}=(a+b)^{2}\}}$

${:\implies \quad \bf \therefore \quad \underline{\underline{\{x-(-11)\}^{2}+\{y-(-6)\}^{2}=(18)^{2}}}}$

On comparing this with the standard equation , we got our centre at (-11,-6) and radius is 18 units